# Ontologies in physics—7: To understand QM, you have to first solve yet another problem with the EM

In the last post, I mentioned the difficulty introduced by (i) the higher-dimensional nature of $\Psi$, and (ii) the definition of the electrostatic $V$ on the separation vectors rather than position vectors.

Turns out that while writing the next post to address the issue, I spotted yet another issue. Its maths is straight-forward (the X–XII standard one). But its ontology is not at all so easy to figure out. So, let me mention it. (This upsets the entire planning I had in mind for QM, and needs small but extensive revisions all over the earlier published EM ontology as well. Anyway, read on.)

In QM, we do use a classical EM quantity, namely, the electrostatic potential energy field. It turns out that the understanding of EM which we have so painfully developed over some 4 very long posts, still may not be adequate enough.

To repeat, the overall maths remains the same. But it’s the physics—rather, the detailed ontological description—which has to be changed. And with it, some small mathematical details would change as well.

I will mention the problem here in this post, but not the solution I have in mind; else this post will become huuuuuge. (Explaining the problem itself is going to take a bit.)

1. Potential energy function for hydrogen atom:

Consider a hydrogen atom in an otherwise “empty” infinite space, as our quantum system.

The proton and the electron interact with each other. To spare me too much typing, let’s approximate the proton as being fixed in space, say at the origin, and let’s also assume a $1D$ atom.

The Coulomb force by the proton (particle 1) on the electron (particle 2) is given by $\vec{f}_{12} = \dfrac{(e)(-e)}{r^2} \hat{r}_{12}$. (I have rescaled the equations to make the constants “disappear” from the equation, though physically they are there, through the multiplication by $1$ in appropriate units.) The potential energy of the electron due to the proton is given by: $V_{12} = \dfrac{(e)(-e)}{r}$. (There is no prefactor of $1/2$ because the proton is fixed.)

The potential energy profile here is in the form of a well, vaguely looking like the letter `V’;  it goes infinitely deep at the origin (at the proton’s position), and its wings asymptotically approach zero at $\pm \infty$.

If you draw a graph, the electron will occupy a point-position at one and only one point on the $r$-axis at any instant; it won’t go all over the space. Remember, the graph is for $V$, which is expressed using the classical law of Coulomb’s.

2. QM requires the entire $V$ function at every instant:

In QM, the measured position of the electron could be anywhere; it is given using Born’s rule on the wavefunction $\Psi(r,t)$.

So, we have two notions of positions for the supposedly same existent: the electron.

One notion refers to the classical point-position. We use this notion even in QM calculations, else we could not get to the $V$ function. In the classical view, the electronic position can be variable; it can go over the entire infinite domain; but it must refer to one and only one point at any given instant.

The measured position of the electron refers to the $\Psi$, which is a function of all infinite space. The Schrodinger evolution occurs at all points of space at any instant. So, the electron’s measured position could be found at any location—anywhere in the infinite space. Once measured, the position “comes down” to a single point. But before measurement, $\Psi$ sure is (nonuniformly) spread all over the infinite domain.

Schrodinger’s solution for the hydrogen atom uses $\Psi$ as the unknown variable, and $V$ as a known variable. Given the form of this equation, you have no choice but to consider the entire graph of the potential energy ($V$) function into account at every instant in the Schrodinger evolution. Any eigenvalue problem of any operator requires the entire function of $V$; a single value at a time won’t do.

Just a point-value of $V$ at the instantaneous position of the classical electron simply won’t do—you couldn’t solve Schrodinger’s equation then.

If we have to bring the $\Psi$ from its Platonic “heaven” to our $1D$ space, we have to treat the entire graph of $V$ as a physically existing (infinitely spread) field. Only then could we possibly say that $\Psi$ too is a $1D$ field. (Even if you don’t have this motivation, read on, anyway. You are bound to find something interesting.)

Now an issue arises.

3. In Coulomb’s law, there is only one value for $V$ at any instant:

The proton is fixed. So, the electron must be movable—else, despite being a point-particle, it is hard to think of a mechanism which can generate the whole $V$ graph for its local potential energies.

But if the electron is movable, there is a certain trouble regarding what kind of a kinematics we might ascribe to the electron so that it generates the whole $V$ field required by the Schrodinger equation. Remember, $V$ is the potential energy of the electron, not of proton.

By classical EM, $V$ at any instant must be a point-property, not a field. But Schrodinger’s equation requires a field for $V$.

So, the only imaginable solutions are weird: an infinitely fast electron running all over the domain but lawfully (i.e. following the laws at every definite point). Or something similarly weird.

So, the problem (how to explain how the $V$ function, used in Schrodinger’s equation) still remains.

4. Textbook treatment of EM has fields, but no physics for multiplication by signs:

In the textbook treatment of EM (and I said EM, not QM), the proton does create its own force-field, which remains fixed in space (for a spatially fixed proton). The proton’s $\vec{E}$ field is spread all over the infinite space, at any instant. So, why not exploit this fact? Why not try to get the electron’s $V$ from the proton’s $\vec{E}$?

The potential field (in volt) of a proton is denoted as $V$ in EM texts. So, to avoid confusion with the potential energy function (in joule) of the electron, let’s denote the proton’s potential (in volt) using the symbol $P$.

The potential field $P$ does remain fixed and spread all over the space at any instant, as desired. But the trouble is this:

It is also positive everywhere. Its graph is not a well, it is a peak—infinitely tall peak at the proton’s position, asymptotically approaching zero at $\pm \infty$, and positive (above the zero-line) everywhere.

Therefore, you have to multiply this $P$ field by the negative charge of electron $e$, so that $P$ turns into the required $V$ field of the electron.

But nature does no multiplications—not unless there is a definite physical mechanism to “convert” the quantities appropriately.

For multiplications with signed quantities, a mechanism like the mechanical lever could be handy. One small side goes down; the other big side goes  up but to a different extent; etc. Unfortunately, there is no place for a lever in the EM ontology—it’s all point charges and the “empty” space, which we now call the aether.

Now, if multiplication of constant magnitudes alone were to be a problem, we could have always taken care of it by suitably redefining $P$.

But the trouble caused by the differing sign still remains!

And that’s where the real trouble is. Let me show you how.

If a proton has to have its own $P$ field, then its role has to stay the same regardless of the interactions that the proton enters into. Whether a given proton interacts with an electron (negatively charged), or with another proton (positively charged), the given proton’s own field still has to stay the same at all times, in any system—else it will not be its own field but one of interactions. It also has to remain positive by sign—even if $P$ is rescaled to avoid multiplications.

But if $V$ has to be negative when an electron interacts with it, and if $V$ also has to be positive when another proton interacts with it, then a multiplication by the signs (by $\pm 1$ must occur. You just can’t avoid multiplications.

But there is no mechanism for the multiplications mandated by the sign conversions.

How do we resolve this issue?

5. The weakness of a proposed solution:

Here is one way out that we might think of.

We say that a proton’s $P$ field stays just the same (positive, fixed) at all times. However, when the second particle is positively charged, then it moves away from the proton; when the second particle is negatively charged, then it moves towards the proton. Since $V$ does require a dot product of a force with a displacement vector, and since the displacement vector does change signs in this procedure, the problem seems to have been solved.

So, the proposed solution becomes: the direction of the motion of the forced particle is not determined only by the field (which is always positive here), but also by the polarity of that particle itself. And, it’s a simple change, you might argue. There is some unknown physics to the very abstraction of the point charge itself, you could say, which propels it this way instead of that way, depending on its own sign.

Thus, charges of opposing polarities go in opposite directions while interacting with the same proton. That’s just how charges interact with fields. By definition. You could say that.

What could possibly be wrong with that view?

Well, the wrong thing is this:

If you imagine a classical point-particle of an electron as going towards the proton at a point, then a funny situation ensues while using it in QM.

The arrows depicting the force-field of the proton always point away from it—except for the one distinguished position, viz., that of the electron, where a single arrow would be found pointing towards the proton (following the above suggestion).

So, the action of the point-particle of the electron introduces an infinitely sharp discontinuity in the force-field of the proton, which then must also seep into its $V$ field.

But a discontinuity like that is not basically compatible with Schrodinger’s equation. It will therefore lead to one of the following two consequences:

It might make the solution impossible or ill-defined. I don’t know enough about maths to tell if this could be true.  But what I can tell is this: Even if a solution is possible (including solutions that possibly may be asymptotic, or are approximate but good enough) then the presence of the discontinuity will sure have an impact on the nature of the solution. The calculated $\Psi$ wouldn’t be the same as that for a $V$ without the discontinuity. That’s inevitable.

But why can’t we ignore the classical point-position of the electron? Well, the answer is that in a more general theory which keeps both particles movable, then we have to calculate the proton’s potential energy too. To do that, we have to take the electric potential (in volts) $P$ of the electron, and multiply it by the charge of the proton. The trouble is: The electric potential field of the electron has singularity at its classical position. So, classical positions cannot be dropped out of the calculations. The classical position of a given particle is necessary for calculating the $V$ field of the other particle, and, vice-versa.

In short, to ensure consistency in the two ways of the interaction, we must regard the singularities as still being present where they are.

And with that consideration, essentially, we have once again come back to a repercussion of the idea that the classical electron has a point position, but its potential energy field in the electrostatic interaction with the proton is spread everywhere.

To fulfill our desire of having a $3D$ field for $\Psi$, we have to have a certain kind of a field for $V$. But $V$ should not change its value in just one isolated place, just in order to allow multiplication by $-1$, because doing so introduces a very bad discontinuity. It should remain the same smooth $V$ that we have always seen in the textbooks on QM.

6. The problem statement, in a nutshell:

So, here is the problem statement:

To find a physically realizable way such that: even if we use the classical EM properties of the electron while calculating $V$, and even if the electron is classically a point-particle, its $V$ function (in joules) should still turn out to be negative everywhere—even if the proton has its own potential field ($P$, in volts) that is positive everywhere in the classical EM.

In short, we have to change the way we look at the physics of the EM fields, and then also make the required changes to any maths, as necessary. Without disturbing the routine calculations either in EM or in QM.

Can it be done? Well, I think the answer is “yes.”

7. A personal note:

While I’ve been having some vague sense of there being some issue “to be looked into later on” for quite some time (months, at least), it was only over the last week, especially over the last couple of days (since the publication of the last post), that this problem became really acute. I always used to skip over this ontology/physics issue and go directly over to using the EM maths involved in the QM. I used to think that the ontology of such EM as it is used in the QM, would be pretty easy to explain—at least as compared to the ontology of QM. Looks like despite spending thousands of words (some 4–5 posts with a total of may be 15–20 K words) there still wasn’t enough of a clarity—about EM.

Not if we adopt the principle, which I discovered on the fly, right while in the middle of writing this series, that nature does no multiplications without there being a physical mechanism for it.

Fortunately, the problem did become clear. Clear enough that, I think, I also found a satisfactory enough solution to it too. Right today (on 2019.10.15 evening IST).

Would you like to give it a try? (I anyway need a break. So, take about a week’s time or so, if you wish.)

Bye for now, take care, and see you the next time.

A song I like:

(Hindi) “jaani o jaani”
Singer: Kishore Kumar
Music: Laxmikant-Pyarelal
Lyrics: Anand Bakshi

History:

— Originally published: 2019.10.16 01:21 IST
— One or two typos corrected, section names added, and a few explanatory sentences added inline: 2019.10.17 22:09 IST. Let’s leave this post right in this form.